How to delete list item automatically in python?
8
If i have a list
for example :
num=[1,2,3,4]
and i want to delete
list[0]
every 2 second, so after 2 second list in num is num = [2,3,4] and 2 second after it will be num =[2,3]
how can i do that?
python
edited Dec 27 '18 at 15:40
yatu
6,8061826
asked Dec 27 '18 at 15:32
TiffanyTiffany
556
add a comment |
8
If i have a list
for example :
num=[1,2,3,4]
and i want to delete
list[0]
every 2 second, so after 2 second list in num is num = [2,3,4] and 2 second after it will be num =[2,3]
how can i do that?
python
edited Dec 27 '18 at 15:40
yatu
6,8061826
asked Dec 27 '18 at 15:32
TiffanyTiffany
556
add a comment |
8
8
8
1
If i have a list
for example :
num=[1,2,3,4]
and i want to delete
list[0]
every 2 second, so after 2 second list in num is num = [2,3,4] and 2 second after it will be num =[2,3]
how can i do that?
python
edited Dec 27 '18 at 15:40
yatu
6,8061826
asked Dec 27 '18 at 15:32
TiffanyTiffany
556
If i have a list
for example :
num=[1,2,3,4]
and i want to delete
list[0]
every 2 second, so after 2 second list in num is num = [2,3,4] and 2 second after it will be num =[2,3]
how can i do that?
python
python
edited Dec 27 '18 at 15:40
yatu
6,8061826
asked Dec 27 '18 at 15:32
TiffanyTiffany
556
edited Dec 27 '18 at 15:40
yatu
6,8061826
asked Dec 27 '18 at 15:32
TiffanyTiffany
556
edited Dec 27 '18 at 15:40
yatu
6,8061826
edited Dec 27 '18 at 15:40
yatu
6,8061826
edited Dec 27 '18 at 15:40
yatu
6,8061826
6,8061826
asked Dec 27 '18 at 15:32
TiffanyTiffany
556
asked Dec 27 '18 at 15:32
TiffanyTiffany
556
asked Dec 27 '18 at 15:32
TiffanyTiffany
556
556
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
9
You could do it using time.sleep, and del to remove the first element:
for i in range(len(num)):
time.sleep(2)
del num[0]
answered Dec 27 '18 at 15:35
yatuyatu
6,8061826
Additionally, if you also want to do something with that first value, you could usenum.pop(0)
– KuboMD
Dec 27 '18 at 15:58
1
Also, although performance is probably not an issue here, for a long list you might want to use collections.deque instead of a list: docs.python.org/3.5/library/collections.html#collections.deque
– Gallaecio
Dec 30 '18 at 10:57
add a comment |
4
Another solution is using list.pop(), as follows:
import time
num=[1,2,3,4]
while num:
num.pop(0)
time.sleep(2)
answered Dec 27 '18 at 15:48
SseinSsein
1,0141921
add a comment |
3
This does so, until eventually it runs out of elements to remove.
import time
num = [1, 2, 3, 4]
while num:
time.sleep(2)
del num[0]
print (num)
answered Dec 27 '18 at 15:38
Joshua FoxJoshua Fox
7,92495178
7
Might be tempted to make thatwhile numinstead ofwhile True- that way the loop will exit gracefully (or not attempt to delete from an empty list to start with) when it's empty rather than raise an exception.
– Jon Clements♦
Dec 27 '18 at 15:41
Good idea. Done
– Joshua Fox
Dec 27 '18 at 19:19
add a comment |
0
Try this code !
Use sleep function by importing time file & del command to remove the element from list.
Code :
import time
num = [1,2,3,4]
for i in num:
time.sleep(2)
print("Delete item ", i)
del i
Output :
Delete item 1
Delete item 2
Delete item 3
Delete item 4
answered Dec 27 '18 at 16:02
UsmanUsman
754614
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
9
You could do it using time.sleep, and del to remove the first element:
for i in range(len(num)):
time.sleep(2)
del num[0]
answered Dec 27 '18 at 15:35
yatuyatu
6,8061826
Additionally, if you also want to do something with that first value, you could usenum.pop(0)
– KuboMD
Dec 27 '18 at 15:58
1
Also, although performance is probably not an issue here, for a long list you might want to use collections.deque instead of a list: docs.python.org/3.5/library/collections.html#collections.deque
– Gallaecio
Dec 30 '18 at 10:57
add a comment |
9
You could do it using time.sleep, and del to remove the first element:
for i in range(len(num)):
time.sleep(2)
del num[0]
answered Dec 27 '18 at 15:35
yatuyatu
6,8061826
Additionally, if you also want to do something with that first value, you could usenum.pop(0)
– KuboMD
Dec 27 '18 at 15:58
1
Also, although performance is probably not an issue here, for a long list you might want to use collections.deque instead of a list: docs.python.org/3.5/library/collections.html#collections.deque
– Gallaecio
Dec 30 '18 at 10:57
add a comment |
9
9
9
You could do it using time.sleep, and del to remove the first element:
for i in range(len(num)):
time.sleep(2)
del num[0]
answered Dec 27 '18 at 15:35
yatuyatu
6,8061826
You could do it using time.sleep, and del to remove the first element:
for i in range(len(num)):
time.sleep(2)
del num[0]
answered Dec 27 '18 at 15:35
yatuyatu
6,8061826
answered Dec 27 '18 at 15:35
yatuyatu
6,8061826
answered Dec 27 '18 at 15:35
yatuyatu
6,8061826
answered Dec 27 '18 at 15:35
yatuyatu
6,8061826
6,8061826
Additionally, if you also want to do something with that first value, you could usenum.pop(0)
– KuboMD
Dec 27 '18 at 15:58
1
Also, although performance is probably not an issue here, for a long list you might want to use collections.deque instead of a list: docs.python.org/3.5/library/collections.html#collections.deque
– Gallaecio
Dec 30 '18 at 10:57
add a comment |
Additionally, if you also want to do something with that first value, you could usenum.pop(0)
– KuboMD
Dec 27 '18 at 15:58
1
Also, although performance is probably not an issue here, for a long list you might want to use collections.deque instead of a list: docs.python.org/3.5/library/collections.html#collections.deque
– Gallaecio
Dec 30 '18 at 10:57
Additionally, if you also want to do something with that first value, you could use
num.pop(0)– KuboMD
Dec 27 '18 at 15:58
Additionally, if you also want to do something with that first value, you could use
num.pop(0)– KuboMD
Dec 27 '18 at 15:58
1
1
Also, although performance is probably not an issue here, for a long list you might want to use collections.deque instead of a list: docs.python.org/3.5/library/collections.html#collections.deque
– Gallaecio
Dec 30 '18 at 10:57
Also, although performance is probably not an issue here, for a long list you might want to use collections.deque instead of a list: docs.python.org/3.5/library/collections.html#collections.deque
– Gallaecio
Dec 30 '18 at 10:57
add a comment |
4
Another solution is using list.pop(), as follows:
import time
num=[1,2,3,4]
while num:
num.pop(0)
time.sleep(2)
answered Dec 27 '18 at 15:48
SseinSsein
1,0141921
add a comment |
4
Another solution is using list.pop(), as follows:
import time
num=[1,2,3,4]
while num:
num.pop(0)
time.sleep(2)
answered Dec 27 '18 at 15:48
SseinSsein
1,0141921
add a comment |
4
4
4
Another solution is using list.pop(), as follows:
import time
num=[1,2,3,4]
while num:
num.pop(0)
time.sleep(2)
answered Dec 27 '18 at 15:48
SseinSsein
1,0141921
Another solution is using list.pop(), as follows:
import time
num=[1,2,3,4]
while num:
num.pop(0)
time.sleep(2)
answered Dec 27 '18 at 15:48
SseinSsein
1,0141921
edited Dec 27 '18 at 16:17
answered Dec 27 '18 at 15:48
SseinSsein
1,0141921
answered Dec 27 '18 at 15:48
SseinSsein
1,0141921
answered Dec 27 '18 at 15:48
SseinSsein
1,0141921
1,0141921
add a comment |
add a comment |
3
This does so, until eventually it runs out of elements to remove.
import time
num = [1, 2, 3, 4]
while num:
time.sleep(2)
del num[0]
print (num)
answered Dec 27 '18 at 15:38
Joshua FoxJoshua Fox
7,92495178
7
Might be tempted to make thatwhile numinstead ofwhile True- that way the loop will exit gracefully (or not attempt to delete from an empty list to start with) when it's empty rather than raise an exception.
– Jon Clements♦
Dec 27 '18 at 15:41
Good idea. Done
– Joshua Fox
Dec 27 '18 at 19:19
add a comment |
3
This does so, until eventually it runs out of elements to remove.
import time
num = [1, 2, 3, 4]
while num:
time.sleep(2)
del num[0]
print (num)
answered Dec 27 '18 at 15:38
Joshua FoxJoshua Fox
7,92495178
7
Might be tempted to make thatwhile numinstead ofwhile True- that way the loop will exit gracefully (or not attempt to delete from an empty list to start with) when it's empty rather than raise an exception.
– Jon Clements♦
Dec 27 '18 at 15:41
Good idea. Done
– Joshua Fox
Dec 27 '18 at 19:19
add a comment |
3
3
3
This does so, until eventually it runs out of elements to remove.
import time
num = [1, 2, 3, 4]
while num:
time.sleep(2)
del num[0]
print (num)
answered Dec 27 '18 at 15:38
Joshua FoxJoshua Fox
7,92495178
This does so, until eventually it runs out of elements to remove.
import time
num = [1, 2, 3, 4]
while num:
time.sleep(2)
del num[0]
print (num)
answered Dec 27 '18 at 15:38
Joshua FoxJoshua Fox
7,92495178
edited Dec 27 '18 at 19:19
answered Dec 27 '18 at 15:38
Joshua FoxJoshua Fox
7,92495178
answered Dec 27 '18 at 15:38
Joshua FoxJoshua Fox
7,92495178
answered Dec 27 '18 at 15:38
Joshua FoxJoshua Fox
7,92495178
7,92495178
7
Might be tempted to make thatwhile numinstead ofwhile True- that way the loop will exit gracefully (or not attempt to delete from an empty list to start with) when it's empty rather than raise an exception.
– Jon Clements♦
Dec 27 '18 at 15:41
Good idea. Done
– Joshua Fox
Dec 27 '18 at 19:19
add a comment |
7
Might be tempted to make thatwhile numinstead ofwhile True- that way the loop will exit gracefully (or not attempt to delete from an empty list to start with) when it's empty rather than raise an exception.
– Jon Clements♦
Dec 27 '18 at 15:41
Good idea. Done
– Joshua Fox
Dec 27 '18 at 19:19
7
7
Might be tempted to make that
while num instead of while True - that way the loop will exit gracefully (or not attempt to delete from an empty list to start with) when it's empty rather than raise an exception.– Jon Clements♦
Dec 27 '18 at 15:41
Might be tempted to make that
while num instead of while True - that way the loop will exit gracefully (or not attempt to delete from an empty list to start with) when it's empty rather than raise an exception.– Jon Clements♦
Dec 27 '18 at 15:41
Good idea. Done
– Joshua Fox
Dec 27 '18 at 19:19
Good idea. Done
– Joshua Fox
Dec 27 '18 at 19:19
add a comment |
0
Try this code !
Use sleep function by importing time file & del command to remove the element from list.
Code :
import time
num = [1,2,3,4]
for i in num:
time.sleep(2)
print("Delete item ", i)
del i
Output :
Delete item 1
Delete item 2
Delete item 3
Delete item 4
answered Dec 27 '18 at 16:02
UsmanUsman
754614
add a comment |
0
Try this code !
Use sleep function by importing time file & del command to remove the element from list.
Code :
import time
num = [1,2,3,4]
for i in num:
time.sleep(2)
print("Delete item ", i)
del i
Output :
Delete item 1
Delete item 2
Delete item 3
Delete item 4
answered Dec 27 '18 at 16:02
UsmanUsman
754614
add a comment |
0
0
0
Try this code !
Use sleep function by importing time file & del command to remove the element from list.
Code :
import time
num = [1,2,3,4]
for i in num:
time.sleep(2)
print("Delete item ", i)
del i
Output :
Delete item 1
Delete item 2
Delete item 3
Delete item 4
answered Dec 27 '18 at 16:02
UsmanUsman
754614
Try this code !
Use sleep function by importing time file & del command to remove the element from list.
Code :
import time
num = [1,2,3,4]
for i in num:
time.sleep(2)
print("Delete item ", i)
del i
Output :
Delete item 1
Delete item 2
Delete item 3
Delete item 4
answered Dec 27 '18 at 16:02
UsmanUsman
754614
answered Dec 27 '18 at 16:02
UsmanUsman
754614
answered Dec 27 '18 at 16:02
UsmanUsman
754614
answered Dec 27 '18 at 16:02
UsmanUsman
754614
754614
add a comment |
add a comment |
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